Chapter 12, Problem 122CP

Classify the bonding in each of the following molecules as ionic, polar covalent, or nonpolar covalent.

   $\begin{array}{l}\text{a}.{\text{ H}}_{\text{2}} \text{d}.{\text{ SO}}_{\text{2}} \text{g}.{\text{ CF}}_{\text{4}}\\ \text{b}.{\text{ K}}_{\text{3}}\text{P} \text{e}.\text{ HF} \text{h}.{\text{ K}}_{\text{2}}\text{S}\\ \text{c}.\text{ NaI} \text{f}.{\text{ CCl}}_{\text{4}}\end{array}$

Interpretation Introduction

(a)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, ${\text{H}}_{\text{2}}$ is nonpolar covalent.

Explanation of Solution

The given molecule is ${\text{H}}_{\text{2}}$.

The electronegativity value of hydrogen is $2.1$.

The difference in the electronegativity values for $\text{H}-\text{H}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for H}-\text{H bond}=2.1-2.1\\ =0\end{array}$

The difference in the electronegativity values for $\text{H}-\text{H}$ bond is $0$.

Thus, the bond formed in ${\text{H}}_{\text{2}}$ is a nonpolar covalent bond with zero electronegativity difference.

Interpretation Introduction

(b)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, ${\text{K}}_3\text{P}$ is polar covalent.

Explanation of Solution

The given molecule is ${\text{K}}_3\text{P}$.

The electronegativity value of potassium is $0.8$.

The electronegativity value of phosphorus is $2.1$.

The difference in the electronegativity values for $\text{K}-\text{P}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for K}-\text{P bond}=2.1-0.8\\ =1.3\end{array}$

The difference in the electronegativity values for $\text{K}-\text{P}$ bond is $1.3$.

Thus, the bond formed in ${\text{K}}_3\text{P}$ is a polar covalent bond with electronegativity difference equal to $1.3$.

Interpretation Introduction

(c)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, $\text{NaI}$ is polar covalent.

Explanation of Solution

The given molecule is $\text{NaI}$.

The electronegativity value of sodium is $0.9$.

The electronegativity value of iodine is $2.5$.

The difference in the electronegativity values for $\text{Na}-\text{I}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for Na}-\text{I bond}=2.5-0.9\\ =1.6\end{array}$

The difference in the electronegativity values for $\text{Na}-\text{I}$ bond is $1.6$.

Thus, the bond formed in $\text{NaI}$ is a polar covalent bond with electronegativity difference equal to $1.6$.

Interpretation Introduction

(d)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, ${\text{SO}}_2$ is polar covalent.

Explanation of Solution

The given molecule is ${\text{SO}}_2$.

The electronegativity value of sulfur is $2.5$.

The electronegativity value of oxygen is $3.5$.

The difference in the electronegativity values for $\text{S}-\text{O}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for S}-\text{O bond}=3.5-2.5\\ =1.0\end{array}$

The difference in the electronegativity values for $\text{S}-\text{O}$ bond is $1.0$.

Thus, the bond formed in ${\text{SO}}_2$ is a polar covalent bond with electronegativity difference equal to $1.0$.

Interpretation Introduction

(e)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, $\text{HF}$ is polar covalent.

Explanation of Solution

The given molecule is $\text{HF}$.

The electronegativity value of hydrogen is $2.1$.

The electronegativity value of fluorine is $4.0$.

The difference in the electronegativity values for $\text{H}-\text{F}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for H}-\text{F bond}=4.0-2.1\\ =1.9\end{array}$

The difference in the electronegativity values for $\text{H}-\text{F}$ bond is $1.9$.

Thus, the bond formed in $\text{HF}$ is a polar covalent bond with electronegativity difference equal to $1.9$.

Interpretation Introduction

(f)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, ${\text{CCl}}_4$ is polar covalent.

Explanation of Solution

The given molecule is ${\text{CCl}}_4$.

The electronegativity value of carbon is $2.5$.

The electronegativity value of chlorine is $3.0$.

The difference in the electronegativity values for $\text{C}-\text{Cl}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for C}-\text{Cl bond}=3.0-2.5\\ =0.5\end{array}$

The difference in the electronegativity values for $\text{C}-\text{Cl}$ bond is $0.5$.

Thus, the bond formed in ${\text{CCl}}_4$ is a polar covalent bond with electronegativity difference equal to $0.5$.

Interpretation Introduction

(g)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, ${\text{CF}}_4$ is polar covalent.

Explanation of Solution

The given molecule is ${\text{CF}}_4$.

The electronegativity value of carbon is $2.5$.

The electronegativity value of fluorine is $4.0$.

The difference in the electronegativity values for $\text{C}-\text{F}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for C}-\text{F bond}=4.0-2.5\\ =1.5\end{array}$

The difference in the electronegativity values for $\text{C}-\text{F}$ bond is $0.5$.

Thus, the bond formed in ${\text{CF}}_4$ is a polar covalent bond with electronegativity difference equal to $1.5$.

Interpretation Introduction

(h)

Interpretation:

The classification of bonding in the given molecule as ionic, polar or nonpolar covalent is to be stated.

Concept Introduction:

The chemical bond that exists between two atoms in which electrons are distributed in an unequal amount is known as polar bond. Due to this unequal distribution of electrons, the molecule will possess the dipole moment in which one end will have partial positive charge and other end will have partial negative charge.

The bonding is formed by the complete transfer of the electrons between the atoms to achieve the stable electron configuration is known as ionic bonding.

The bonding is formed by the sharing of the electrons between the atoms to achieve the stable electron configuration is known as covalent bonding.

The ability that is possessed by an atom to attract the bonding pair of electrons towards itself is known as electronegativity. The electronegativity decreases while going down the group and increases while moving left to right in the period.

Answer to Problem 122CP

The bonding present in the given molecule, ${\text{K}}_{\text{2}}\text{S}$ is polar covalent.

Explanation of Solution

The given molecule is ${\text{K}}_{\text{2}}\text{S}$.

The electronegativity value of potassium is $0.8$.

The electronegativity value of sulfur is $2.5$.

The difference in the electronegativity values for $\text{K}-\text{S}$ bond is,

$\begin{array}{c}\text{Difference}\text{in}\text{electronegativity for K}-\text{S bond}=2.5-0.8\\ =1.7\end{array}$

The difference in the electronegativity values for $\text{K}-\text{S}$ bond is $1.7$.

Thus, the bond formed in ${\text{K}}_{\text{2}}\text{S}$ is a polar covalent bond with electronegativity difference equal to $1.7$.